--- title: "Critical frequency and skip distance (flat earth)" subject: "Propagation and Antenna" chapter: 4 key: "semester_6/propagation_and_antenna:critical-frequency-skip-distance-flat" source: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/critical-frequency-skip-distance-flat site: BCT Engineering --- # Critical frequency and skip distance (flat earth) Propagation and Antenna · Chapter 4: Propagation and Radio Frequency Spectrum - Asked in: 2080 Asoj, 7 marks; 2072 Magh, 8 marks (asked 2 times) - Video: [Skip Distance (Definition, Basics, Derivation, Formula & Calculation) Explained](https://www.youtube.com/watch?v=JoV6IAyOxEA) — Engineering Funda - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/propagation_and_antenna/critical-frequency-skip-distance-flat - Whole chapter: https://www.bctengineer.com/md/semester_6/propagation_and_antenna/chapter-4 ## Question Find out the relation between the critical frequency (fcr) and skip distance (D) considering that the Earth is flat for both antennas. ## Short answer D = 2h sqrt((f/fcr)^2 - 1) and fcr = 2hf / sqrt(4h^2 + D^2), where h is virtual height and f is operating frequency. At f = fcr, skip distance is zero. ## Answer **Critical frequency (fcr)** is the highest frequency reflected by a layer at vertical incidence, fcr = 9√Nmax. **Skip distance (D)** is the minimum ground distance at which a sky wave of frequency *f* (> fcr) returns to earth. The relation follows from the refraction condition and flat-earth geometry. **Assumptions:** flat earth and flat ionosphere; reflection at virtual height *h*; transmitter and receiver both on the ground (heights neglected); symmetric path. ```text ionosphere (virtual height h) ~~~~~~~~~~~~~~~~~~ P ~~~~~~~~~~~~~~~~~~ / | \ /θi| \ / |h \ / | \ flat earth T-----M-----R <--- D ----> TM = MR = D/2, θi = angle of incidence ``` Refractive index of an ionised layer (electron density N per m³, frequency f in Hz): ```text n = √(1 − 81N / f²) ``` By Snell's law, a wave entering at angle θi turns back where the ray becomes horizontal (refraction angle 90°): ```text sin θi = n = √(1 − 81Nmax / f²) cos² θi = 81Nmax / f² f = 9√Nmax / cos θi ``` For vertical incidence (θi = 0) this gives the critical frequency fc = 9√Nmax, so ```text f_MUF = fc sec θi (secant law) ``` From the flat-earth geometry (triangle TMP): ```text tan θi = (D/2) / h sec θi = √(1 + tan² θi) = √(1 + D² / 4h²) f_MUF = fc √(1 + D² / 4h²) = fc √(4h² + D²) / (2h) ``` Solving for the skip distance: ```text (f_MUF / fc)² = 1 + D² / 4h² D = 2h √[(f_MUF / fc)² − 1] ``` Here f_MUF is the working frequency *f* for which D is the skip distance. So the required relation is: ```text D = 2h √[(f / fcr)² − 1] or fcr = f / √(1 + D² / 4h²) = 2hf / √(4h² + D²) ``` **Observations** - If f = fcr, D = 0: the wave returns even straight overhead, so there is no skip zone. - For f > fcr, D increases with f; a higher frequency needs a more oblique ray to return. - A higher layer (bigger *h*) gives a larger skip distance for the same f/fcr ratio. - At night fcr falls (lower Nmax), so the skip distance for the same frequency increases; operators lower the frequency at night. **Example:** h = 250 km, fcr = 6 MHz, f = 12 MHz: D = 2 × 250 × √(2² − 1) = 500 × 1.732 = 866 km. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which relation links the usable frequency with critical frequency? A. f = fc sec θi B. f = fc sin θi C. f = fc tan θi D. f = fc / sec² θi Answer: A. From cos²θi = 81Nmax/f², f = 9√Nmax sec θi = fc sec θi. (Key point: Secant law) 2. Which relation between fcr and D is correct? A. D = 2h √[(f/fcr)² − 1] B. D = 2h √[(f/fcr)² + 1] C. D = h (f/fcr) D. D = 2h (fcr/f) Answer: A. Using sec θi = √(1 + D²/4h²) and f = fcr sec θi, D = 2h√[(f/fcr)² − 1]. (Key point: Skip distance relation) 3. What is D when f = fcr? A. Infinite B. Zero, so there is no skip zone C. 2h D. Equal to h Answer: B. At f = fcr the bracket is zero, so the wave returns even overhead. (Key point: Behaviour at f = fcr) 4. For h = 250 km, fcr = 6 MHz, f = 12 MHz, D = ? and what happens at night when fcr falls? A. 866 km; skip distance increases for the same frequency B. 500 km; skip distance decreases C. 1500 km; skip distance is unchanged D. 866 km; skip distance decreases Answer: A. D = 2 x 250 x √(4 − 1) = 866 km. Lower fcr raises f/fcr, so D increases and operators lower the frequency at night. 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